1. Set up motion of the first ball

Let the first ball be thrown at time $t=0$ with speed $u=35\,\text{m/s}$ upward.

Its height after time $t$ is $y_1 = ut - \frac{1}{2}gt^2 = 35t - 5t^2.$

2. Set up motion of the second ball

The second ball is thrown upward after $3\,\text{s}$ with the same speed.

So for $t\ge 3$, its height is $y_2 = 35(t-3) - 5(t-3)^2.$

3. Condition for collision of the balls

Since each ball has radius $5\,\text{cm} = 0.05\,\text{m}$, they collide when the distance between their centers becomes $0.05+0.05 = 0.10\,\text{m}.$

The first ball is always above the second one until collision, so $y_1 - y_2 = 0.10.$

Now, $y_1 - y_2 = (35t-5t^2) - \left[35(t-3)-5(t-3)^2\right].$

Simplify: $35(t-3)=35t-105,$ $(t-3)^2=t^2-6t+9,$ $-5(t-3)^2=-5t^2+30t-45.$

Thus, $y_2 = 35t-105-5t^2+30t-45 = -5t^2+65t-150.$

Hence, $y_1-y_2=(35t-5t^2)-(-5t^2+65t-150)=150-30t.$

Set this equal to $0.10$: $150-30t=0.10$ $30t=149.9$ $t=\frac{149.9}{30}=4.996\overline{6}\,\text{s}.$

This is the time after the first ball was thrown.

4. Find the height of collision

Use height of the first ball: $y = 35t-5t^2.$

Substitute $t=\frac{149.9}{30}$: $y \approx 35(4.9967)-5(4.9967)^2.$

$y \approx 174.8833 - 124.8334 = 50.0499\,\text{m}.$

So the collision occurs at approximately $\boxed{50\,\text{m}}.$

5. Check against stored answer

Stored correct answer: $50$

Derived answer: $50$

They match.